Rewrite the function by completing the square. $h(x)= x^{2} +3 x -18$ $h(x)=$
Explanation: $\begin{aligned} h(x)&= x^2 +3 x -18 \\\\ &= \left(x^2 +3 x\right) -18 \end{aligned}$ Now we want to complete $x^2 +3 x$ into a perfect square. To do that, we should add $\left(\dfrac{{3}}{2}\right)^2={\dfrac{9}{4}}$ to it: $x^2{+3}x+{\dfrac{9}{4}}=\left(x +\dfrac{3}{2}\right)^2$ We add ${\dfrac{9}{4}}$ inside the parentheses, and subtract ${1}\cdot{\dfrac{9}{4}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=} \left(x^2 +3 x\right) -18 \\\\ &=\left(x^2 +3 x+{\dfrac{9}{4}}\right) -18 -{1}\cdot{\dfrac{9}{4}} \\\\ &= \left(x +\dfrac{3}{2}\right)^2 -18 -\dfrac{9}{4} \\\\ &= \left(x +\dfrac{3}{2}\right)^2 -\dfrac{81}{4} \end{aligned}$ In conclusion, the function after completing the square is written as: $h(x)= \left(x +\dfrac{3}{2}\right)^2 -\dfrac{81}{4}$